Question
Prove the following.
$1+\frac{\cot^2\alpha}{1+\text{cosec}\alpha}=\text{cosec}\alpha$
$1+\frac{\cot^2\alpha}{1+\text{cosec}\alpha}=\text{cosec}\alpha$
✓
Answer
LHS $=1+\frac{\cot^2\alpha}{1+\text{cosec}\alpha}=1+\frac{\cos^2\alpha/\sin^2\alpha}{1+1/\sin\alpha}$ $\bigg[\because\ \cot\theta=\frac{\cos\theta}{\sin\theta}\text{ and cosec}\theta=\frac{1}{\sin\theta}\bigg]$
$=1+\frac{\cos^2\alpha}{\sin\alpha(1+\sin\alpha)}=\frac{\sin\alpha(1+\sin\alpha)+\cos^2\alpha}{\sin\alpha(1+\sin\alpha)}$
$=\frac{\sin\alpha+(\sin^2\alpha+\cos^2\alpha)}{\sin\alpha(1+\sin\alpha)}\ [\because\ \sin^2\theta+cos^2\theta=1]$
$=\frac{(\sin\alpha+1)}{\sin\alpha(\sin\alpha+1)}=\frac{1}{\sin\alpha}\ \Big[\because\ \text{cosec}\theta=\frac{1}{\sin\theta}\Big]$
$=\text{cosec}\alpha=$ RHS
$=1+\frac{\cos^2\alpha}{\sin\alpha(1+\sin\alpha)}=\frac{\sin\alpha(1+\sin\alpha)+\cos^2\alpha}{\sin\alpha(1+\sin\alpha)}$
$=\frac{\sin\alpha+(\sin^2\alpha+\cos^2\alpha)}{\sin\alpha(1+\sin\alpha)}\ [\because\ \sin^2\theta+cos^2\theta=1]$
$=\frac{(\sin\alpha+1)}{\sin\alpha(\sin\alpha+1)}=\frac{1}{\sin\alpha}\ \Big[\because\ \text{cosec}\theta=\frac{1}{\sin\theta}\Big]$
$=\text{cosec}\alpha=$ RHS
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